Module 5 : MODERN PHYSICS
Lecture 27 : Nuclear Energy
  Example 20
  Calculate the energy released in the fission reaction
$\displaystyle ^{235}_{92}U + n \longrightarrow ^{93}_{37}Rb + ^{141}_{55}Cs + 2n$
  Solution :
The masses of the reactants and the product (in units of u)are as follows :
  Reactants :
 
mass of $ ^{235}_{92}$U = 235.043929
mass of n = 1.008665
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Total mass of reactants = 236.052594
  Products :
 
mass of $ ^{93}_{37}$Rb = 92.922042
mass of $ ^{141}_{55}$Cs = 140.920046
mass of 2 n = 2.017330
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Total mass of product = 235.859418
  The products being lighter than the reactants by $ 236.052594-235.859418= 0.193176$ u. This amounts to $ 0.193176\times 931.5=179.94$ MeV of energy.
 
Fission produces about 200 Mev of energy of which about 175 MeV is the kinetic energy of fission fragments, the reamaining energy is distributed as the kinetic energy of neutrons and energy associated with photons, neutrinos and other radioactive products.

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